Animated Solution for Physics - Oscillations: The mass $M$ shown in the figure oscillates in simple harmonic motion with amplitude $A$. The amplitude of the point $P$ is

The Magic of Coupled Oscillations

In the vast universe of classical mechanics, few systems are as elegant and ubiquitous as the simple harmonic oscillator. From the microscopic vibrations of atoms in a crystal lattice to the macroscopic sway of skyscrapers in an earthquake, the mathematics of simple harmonic motion (SHM) governs them all. Today, we embark on a journey to deconstruct a classic problem from the prestigious JEE Advanced exam: a block of mass $M$ coupled to a rigid wall through two springs connected in series.

Series Springs

Imagine a block of mass $M$ resting on a frictionless horizontal surface. It is connected to a wall, but not by a single spring. Instead, it is linked via a chain of two springs: the first with spring constant $k_1$, and the second with spring constant $k_2$. The point where these two springs meet is labeled $P$.

When the block is pulled and released, it executes SHM with an amplitude $A$. But what happens to the junction point $P$? Does it remain stationary? No, it must move as well! Our mission is to find the amplitude of this point $P$.

A Physics Secret

The key to unlocking this problem lies in a fundamental, yet often overlooked, principle of mechanics: the massless junction constraint.

The point $P$ is a junction between two springs, and it has no mass of its own ($m_P=0$). According to Newton's second law, the net force on any object is equal to its mass times its acceleration:

If there were any net force on point $P$, its acceleration would have to be infinite, which is physically impossible. Therefore, the net force on point $P$ must be exactly zero at every single instant of time!

This means the pulling force exerted by the first spring on point $P$ must be perfectly balanced by the pulling force exerted by the second spring:

Using Hooke's Law, we can write this as:

where $x_1$ is the extension of the first spring, and $x_2$ is the extension of the second spring.

Hooke's Law in Action

Now, let's relate these individual extensions to the total displacement of the block $M$.

The total displacement $x$ of the block from its equilibrium position is simply the sum of the extensions of both springs:

Since the first spring is attached directly to the wall, the displacement of point $P$ from its equilibrium position is exactly equal to the extension of the first spring, $x_1$.

Therefore, the amplitude of point $P$ (its maximum displacement) is simply the maximum value of $x_1$.

Let's express $x_2$ in terms of $x_1$ using our force balance equation:

Substituting this into the total displacement equation, we get:

Let's factor out $x_1$ to simplify the expression:

Now, we can solve for $x_1$ in terms of the total displacement $x$:

Since the maximum displacement of the block $M$ is its amplitude $A$, the maximum displacement of point $P$ (its amplitude $A_P$) occurs when $x=A$:

This is a beautiful and intuitive result! It shows that the displacement is shared between the two springs in inverse proportion to their stiffness. The softer spring will stretch more, while the stiffer spring will stretch less.

The Equivalent Spring Method

To truly master this concept, let's look at it from another perspective: the equivalent spring constant method.

Since the two springs are in series, their equivalent spring constant $k_{\text{eq}}$ is given by:

At the maximum displacement $A$, the maximum restoring force acting on the block (and thus transmitted through the entire series chain) is:

Since this same force acts on the first spring, we can write:

Equating the two expressions:

Canceling $k_1$ from both sides, we get:

Both methods lead us to the exact same elegant conclusion! This is the hallmark of consistent physical laws.